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Part of what it means to be a functor between two categories is to have a map of morphisms e.g. $F$ sends $f: A \to B$ to $Ff: FA \to FB$.

Suppose $F$ is a functor from a category to itself, and that category has (some or all) exponential objects. $B^A$ intuitively corresponds to an object of morphisms $A \to B$, so I'd hope to be able to apply the morphism-mapping of the functor $F$ to get $FB^{FA}$. I can do this in $\mathbf{Set}$ in an obvious way, but I can't see how to frame that idea category-theoretically so that I might generalise it. So my questions are:

  • Under what conditions can I "internalise" the morphism map of $F$ to get a morphism $B^A \to FB^{FA}$?
  • If that doesn't work, what if I replace $B^A$ with the hom-functor of a category enriched over itself, or some other internal notion of function?
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up vote 5 down vote accepted

This difficulty is exactly the reason why enriched category theory isn't vacuous. There is a result of Kock which spells out necessary and sufficient conditions for a functor to have an enrichment – without ever mentioning internal homs! Here it is.

Definition. Let $\mathcal{C}$ be a monoidal closed category (such as a cartesian closed category). A strength for an endofunctor $T : \mathcal{C} \to \mathcal{C}$ is a natural transformation $\sigma_{X, Y} : X \otimes T Y \to T (X \otimes Y)$ satisfying these axioms: $\newcommand{\id}{\textrm{id}}$

  • $T \lambda_X \circ \sigma_{I, X} = \lambda_{T X}$, where $I$ is the monoidal unit and $\lambda_X : I \otimes X \to X$ is the unitor.

  • $\sigma_{X, Y \otimes Z} \circ (\id_X \otimes \sigma_{Y,Z}) \circ \alpha_{X,Y,T Z} = T \alpha_{X,Y,Z} \circ \sigma_{X \otimes Y, Z}$, where $\alpha_{X,Y,Z} : (X \otimes Y) \otimes Z \to X \otimes (Y \otimes Z)$ is the associator.

Theorem. There is a bijection between natural transformations of type $$\sigma_{X,Y} : X \otimes T Y \to T (X \otimes Y)$$ and natural transformations of type $$t_{X,Y} : \mathcal{H}om (X, Y) \to \mathcal{H}om (T X, T Y)$$ such that $\sigma_{X,Y}$ is a strength for $T$ if and only if $t$ is an enrichment for $T$.

Proof. I briefly indicate where the bijection comes from; the details can be found in Theorem 3.2.17 of my notes. Suppose $\sigma$ is given. To construct $t$, it is enough to construct morphisms $\mathcal{H}om (X, Y) \otimes T X \to T Y$ satisfying an appropriate extranaturality condition. But there is a fairly obvious one we can use, namely $T (\textrm{ev}_{X, Y}) \circ \sigma_{\mathcal{H}om (X, Y), X}$. Conversely, given $t$, we construct $\sigma$ as the tensor–hom transpose of $t_{Y, X \otimes Y} \circ D_{X, Y}$, where $D_{X, Y} : X \to \mathcal{H}om (Y, X \otimes Y)$ is the unit of the tensor–hom adjunction.

Ultimately, however, the idea comes from the following observation: ordinary functors between cocomplete categories have a canonical comparison map between copowers (by the universal property thereof), and the same is true for enriched functors between cocomplete enriched categories, where "copower" is replaced by "tensor product". This extra structure tells us how to handle "non-discrete" families of morphisms in enriched categories.

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Some thoughts: If $F$ preserves products we can apply it to the evaluation map $e:B^{A}\times A\rightarrow B$ to get a map $F(e):F(B^{A})\times F(A)\rightarrow F(B)$, which, if $F(A)$ is also exponentiable, gives us a map $\bar{F(e)}:F(B^{A})\rightarrow F(B)^{F(A)}$ by transposition. In general, I can't think of a reason that there should be any useful morphism $B^{A}\rightarrow F(B^{A})$.

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